Science:Math Exam Resources/Courses/MATH110/April 2011/Question 04 (b)

MATH110 April 2011

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Question 04 (b)
Given a curve, its curvature K is a measure of how much it deviates from a straight line. It can be computed using the following formula: $K={\frac {y''}{\left(1+(y')^{2}\right)^{3/2}}}$ In this question you will calculate the curvature of the curve x4 + y4 = 2 at a certain point. Use your answer from part (a) to calculate $\displaystyle y''$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Again, use implicit differentiation. You will need to use the quotient rule as well.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. To find $y''$ we simply need to differentiate the expression $y'=-x^{3}/y^{3}$ from part (a). Using implicit differentiation and the quotient rule, we have: ${\begin{aligned}y''&={\frac {-3x^{2}(y^{3})-(-x^{3})3y^{2}y'}{(y^{3})^{2}}}\\&={\frac {3x^{2}y^{2}(-y+xy')}{y^{6}}}\\&={\frac {3x^{2}(-y+xy')}{y^{4}}}\\\end{aligned}}$

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